The equation (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.
The horizontal momentum is given by:
Where:
- m₁: is the mass of the lab cart = 15 kg
- m₂: is the <em>mass </em>of the object dropped = 2 kg
- : is the initial velocity of the<em> lab cart </em>
- : is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)
- : is the final velocity of the<em> lab cart </em>
- : is the <em>final velocity</em> of the <em>object </em>
Then, the horizontal momentum is:
When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:
Therefore, the equation represents the horizontal momentum (option 3).
Learn more about linear momentum here:
I hope it helps you!
Answer:
Explanation:
The force of attraction between 2 masses.
Top of the U ramp: potential energy is the highest, while kinetic energy is the lowest
Bottom of the U ramp(aka the curve part): potential energy is the lowest and the kinetic energy is the highest
THEREFORE, PE and KE have an INVERSE RELATIONSHIP.
Wow ! I understand your shock. I shook and vibrated a little
when I looked at this one too.
The reason for our shock is all the extra junk in the question,
put there just to shock and distract us.
"Neutron star", "5.5 solar masses", "condensed burned-out star".
That's all very picturesque, and it excites cosmic fantasies in
out brains when we read it, but it's just malicious decoration.
It only gets in the way, and doesn't help a bit.
The real question is:
What is the acceleration of gravity 2000 m from
the center of a mass of 1.1 x 10³¹ kg ?
Acceleration of gravity is
G · M / R²
= (6.67 x 10⁻¹¹ N·m²/kg²) · (1.1 x 10³¹ kg) / (2000 m)²
= (6.67 x 10⁻¹¹ · 1.1 x 10³¹ / 4 x 10⁶) (N) · m² · kg / kg² · m²
= 1.83 x 10¹⁴ (kg · m / s²) · m² · kg / kg² · m²
= 1.83 x 10¹⁴ m / s²
That's about 1.87 x 10¹³ times the acceleration of gravity on
Earth's surface.
In other words, if I were standing on the surface of that neutron star,
I would weigh 1.82 x 10¹² tons, give or take.